Let $$M=\begin{bmatrix}A&B\\C&D\end{bmatrix}$$ be a matrix in $\mathbb F_2^{n\times n}$ where $A\in\mathbb F_2$ and $D\in\mathbb F_2^{(n-1)\times(n-1)}$ are square.

Through the determinant result on Schur complement $$\det(M)=\det(D-CB)$$ if $A=1$ holds.

It suggests an algorithm for $\det(M)$. Without loss of generality assume $A=M_{1,1}=1$.

At step $i\in\{0,1,\dots,n-1\}$ we permute the $(n-i)\times(n-i)$ matrix $M_i=\begin{bmatrix}A_i&B_i\\C_i&D_i\end{bmatrix}$ (where $M_0=M$ and $A_i\in\mathbb F_2$ and $D_i\in\mathbb F_2^{(n-i-1)\times(n-i-1)}$ holds) by $P_{1i}M_iP_{2i}$ where $P_{1i},P_{2i}\in\mathbb F_2^{(n-i)\times(n-i)}$ are permutation matrices satisfying $M_i(1,1)=1$ (first entry is $1$ and so invertible) and set $M_{i+1}=D_i-C_iB_i\in\mathbb F_2^{(n-i-1)\times(n-i-1)}$.

At end of $n$ iterations the remaining entry is determinant of $M$.

My question:

Is there a single permutation $P_1,P_2\in\mathbb F_2^{n\times n}$ we can apply to $M$ and set $M_0=P_1MP_2$ so that $P_{1i}=P_{2i}$ are identity matrices in $\mathbb F_2^{(n-i)\times(n-i)}$ at every $i\in\{1,2,\dots,n-1\}$ so that either $M_i$ is a zero matrix at an $i\in\{1,2,\dots,n-1\}$ or $M_{n-1}=1$ holds?

My guess is the answer is no since number of matrices is $2^{\Omega(n^2)}$ while number of permutations is $2^{O(n\log n)}$. We require every permutation to depermute for $2^{\Omega(n^2)}$ matrices which is unlikely.